Let $f(x)=\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}e^{\frac{x}{n}}$. Let $f(x)=\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}e^{\frac{x}{n}}$.
Is f differentiable on $\mathbb{R}$?
I can prove  the series does not uniformly convergent on $\mathbb{R}$. But this does not imply f is not differentiable. 
 A: Write 
$$ f(x) - f(0) = \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}} \left( e^{x/n} - 1 \right) $$
Now $e^{x/n} - 1 = x/n + O((x/n)^2)$, and since $\sum_n 1/n^{3/2}$ converges we find that this series converges uniformly on compact subsets of $\mathbb C$.  Thus $f(x) - f(0)$ is an entire function, and then so is $f(x)$.
A: 
I thought it might be instructive to present a "brute-force" way forward that relies on straightforward analysis and an inequality for the exponential function.  To that end, we now proceed.


$$\begin{align}
\left|\frac{f(x+h)-f(x)}h -\sum_{n=1}^\infty\frac{(-1)^n}{n^{3/2}}e^{x/n}\right|&=\left|\sum_{n=1}^\infty \frac{(-1)^n}{n^{1/2}}e^{x/n}\left(\frac{e^{h/n}-1}{h}-\frac1n\right)\right|\\\\
&\le \sum_{n=1}^\infty \frac{1}{n^{1/2}}e^{x/n}\left|\frac{e^{h/n}-1}{h}-\frac1n\right| \tag 1
\end{align}$$
Then, from the upper bound for $e^x$ established in THIS ANSWER, we have
$$\left|\frac{e^{h/n}-1}{h}-\frac1n\right|
\le \frac{h}{n(n-h)} \tag 2$$
for $h<n$.  Taking $h<1$, we find using $(2)$ in $(1)$
$$\begin{align}
\left|\frac{f(x+h)-f(x)}h -\sum_{n=1}^\infty\frac{(-1)^n}{n^{3/2}}e^{x/n}\right|&\le h\left(\sum_{n=1}^\infty \frac{1}{n^{1/2}}e^{x/n}\frac{1}{n(n-1)}\right)\tag 3
\end{align}$$
Inasmuch as the right-hand side of $(3)$ approaches $0$ as $h\to 0$, we conclude that

$$\bbox[5px,border:2px solid #C0A000]{f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\sum_{n=1}^\infty\frac{(-1)^n}{n^{3/2}}e^{x/n}}$$  

And we are done!
